If p, q, r are in A.P. and x, y, z are in G.P., then $$x^{\,q-r} \, y^{\,r-p} \, z^{\,p-q}$$​ is:

Correct option is B

Given:
p, q, r are in A.P.
x, y, z are in G.P.
Formula used:
If p, q, r are in A.P. then
​$$2q = p + r$$​​
If x, y, z are in G.P. then
​$$y^{2} = xz$$​​
Solution:
Since p, q, r are in A.P.:
​$$q − r = p − q\\r − p = 2(q − p)$$​​
Given expression:
​$$x^{q-r} y^{r-p} z^{p-q}$$​​
Group the terms:
​$$= x^{q-r} z^{p-q} \cdot y^{r-p}= \left(\frac{x}{z}\right)^{q-r} \cdot y^{r-p}$$​​
But from G.P., $$y^{2} = xz$$​​
So, $$\frac{x}{z} = \frac{y^{2}}{z^{2}} = \left(\frac{y}{z}\right)^{2}$$​​
Using the A.P. relation, the total exponent of y becomes zero.
Hence,
​$$x^{q-r} y^{r-p} z^{p-q} = 1$$​​
The correct answer is (b) 1.